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Question

Let A and B be any two points on the lines represented by 4x29y2=0. If the area of triangle OAB is 5 (O is origin) then which of the following is the possible equation of the locus of midpoint of AB ? 
  1. 4x2+9y2+30=0
  2. 4x29y230=0
  3. 9x24y230=0
  4. 9x2+4y230=0


Solution

The correct option is B 4x29y230=0

Area of OAB=12OA.OBsinθ 
                      =5

OA=133h ; OB=133k

5=12.133h.133k.sinθ

tanθ=23(23)1+23(23)=125sinθ=1213

|hk|=152hk=±152

Midpoint of AB=⎜ ⎜ ⎜h+k2,2h32k32⎟ ⎟ ⎟

     =(h+k2,hk3)
h+k=2x;  hk=3y
2h=2x+3y
2k=2x3y
4hk=4x29y2
4x29y2=±30

Alternate Solution:
Area of OAB=12∣ ∣ ∣ ∣ ∣10 01k2k31h 2h3∣ ∣ ∣ ∣ ∣
5=122kh3+2kh3
|hk|=152hk=±152

Midpoint of AB=⎜ ⎜ ⎜h+k2,2h32k32⎟ ⎟ ⎟

     =(h+k2,hk3)
h+k=2x;  hk=3y
2h=2x+3y
2k=2x3y
4hk=4x29y2
4x29y2=±30

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